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Question: In how many ways can five people be distributed in three different rooms if no room must be empty?...

In how many ways can five people be distributed in three different rooms if no room must be empty?

Explanation

Solution

In this problem we have to find a number of ways to distribute the given number of persons into a given number of rooms with the condition in given. In order to solve this question we have to apply the concept of permutation since it is a question of arrangement in an order. By substituting the given values with the given condition we will get the required result.

Complete step-by-step answer:
It is given in the question that the total number of persons present=5 = 5
Further it is stated that we need to arrange them in 33 different rooms leaving no room empty. So from the question it is clear we have to arrange 55 people in 33 different rooms.
Therefore we can write that-
Total number of ways to arrange them == Number of ways in which groups can be formed ×\times Number of ways groups can be arranged in different rooms
⇒[5!3!×12!]3!+[5!(2!)2×12!]×3!\Rightarrow \left[ {\dfrac{{5!}}{{3!}} \times \dfrac{1}{{2!}}} \right]3! + \left[ {\dfrac{{5!}}{{{{(2!)}^2}}} \times \dfrac{1}{{2!}}} \right] \times 3!
Since we have to arrange 55 people in 33 rooms.
⇒[5×4×3!3!×12!](3×2!)+[5×4×3×2×1(2×1)2×12×1]×3×2×1\Rightarrow \left[ {\dfrac{{5 \times 4 \times 3!}}{{3!}} \times \dfrac{1}{{2!}}} \right]\left( {3 \times 2!} \right) + \left[ {\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{{{(2 \times 1)}^2}}} \times \dfrac{1}{{2 \times 1}}} \right] \times 3 \times 2 \times 1
Expanding the factorial terms to solve,
⇒[5×4×3]+[5×4×3×2×1(2)2×12×1]×3×2×1\Rightarrow \left[ {5 \times 4 \times 3} \right] + \left[ {\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{{{(2)}^2}}} \times \dfrac{1}{{2 \times 1}}} \right] \times 3 \times 2 \times 1
Cancelling the similar terms to get the result,
⇒[5×4×3]+[5×4×34]×3×2×1\Rightarrow \left[ {5 \times 4 \times 3} \right] + \left[ {\dfrac{{5 \times 4 \times 3}}{4}} \right] \times 3 \times 2 \times 1
Hence we get,
⇒[5×4×3]+[5×3]×3×2×1\Rightarrow \left[ {5 \times 4 \times 3} \right] + \left[ {5 \times 3} \right] \times 3 \times 2 \times 1
⇒60+90=150\Rightarrow 60 + 90 = 150

∴\therefore Thus the number of ways to arrange them is 150150

Note: Permutation can be defined as the technique of arranging all the members in a group. In other ways, we can say that if the set has already been ordered then re-ordering the elements is called the process of permuting. It is useful almost in every part of mathematics. It is basically used for arranging people, digits, numbers etc.
The basic formula for permutation npr=n!(n−r)!{}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}